Exercise 2.2 Solution Example - Hoff, A First Course in Bayesian Statistical Methods
標準ベイズ統計学 演習問題 2.2 解答例

\begin{align*} \text{E} [a_1 Y_1 + a_2 Y_2] &= a_1 \text{E} [Y_1] + a_2 \text{E} [Y_2] \\ & = a_1 \mu_1 + a_2 \mu_2 \\ \\ \text{Var} [a_1 Y_1 + a_2 Y_2] &= \text{E} [(a_1 Y_1 + a_2 Y_2)^2] - \text{E} [a_1 Y_1 + a_2 Y_2]^2 \\ &= \text{E} [a_1^2 Y_1^2 + a_2^2 Y_2^2 + 2 a_1 a_2 Y_1 Y_2] - \text{E} [a_1 Y_1 + a_2 Y_2]^2 \\ &= a_1^2 \text{E} [Y_1^2] + a_2^2 \text{E} [Y_2^2] + 2 a_1 a_2 \text{E} [Y_1 Y_2] - (a_1 \mu_1 + a_2 \mu_2)^2\\ &= a_1^2 (\text{E} [Y_1^2] - \mu_1^2) + a_2^2 (\text{E} [Y_2^2] - \mu_2^2) + 2 a_1 a_2 \text{E} [Y_1] \text{E} [Y_2] - 2 a_1 a_2 \mu_1 \mu_2 \\ & \qquad \qquad ( \because Y_1 \text{ and } Y_2 \text{ are independent })\\ &= a_1^2 \sigma_1^2 + a_2^2 \sigma_2^2 \end{align*}

\begin{align*} \text{E} [a_1 Y_1 - a_2 Y_2] &= a_1 \text{E} [Y_1] - a_2 \text{E} [Y_2] \\ &= a_1 \mu_1 - a_2 \mu_2 \\ \\ \text{Var} [a_1 Y_1 - a_2 Y_2] &= \text{E} [(a_1 Y_1 - a_2 Y_2)^2] - \text{E} [a_1 Y_1 - a_2 Y_2]^2 \\ &= \text{E} [a_1^2 Y_1^2 + a_2^2 Y_2^2 - 2 a_1 a_2 Y_1 Y_2] - \text{E} [a_1 Y_1 - a_2 Y_2]^2 \\ &= a_1^2 \text{E} [Y_1^2] + a_2^2 \text{E} [Y_2^2] - 2 a_1 a_2 \text{E} [Y_1 Y_2] - (a_1 \mu_1 - a_2 \mu_2)^2\\ &= a_1^2 (\text{E} [Y_1^2] - \mu_1^2) + a_2^2 (\text{E} [Y_2^2] - \mu_2^2) - 2 a_1 a_2 \text{E} [Y_1] \text{E} [Y_2] + 2 a_1 a_2 \mu_1 \mu_2 \\ & \qquad \qquad ( \because Y_1 \text{ and } Y_2 \text{ are independent })\\ &= a_1^2 \sigma_1^2 + a_2^2 \sigma_2^2 \end{align*}